Can you please explain what the difference between an integral curve and the solution of a differential equation is? My book gives an example that $$\frac {dy}{dx}=\frac {y}{x}$$ defines a direction field everywhere except at the origin. the function $y=kx$ is a solution of this equation, and the integral curve is given by $ax+by=0$, where $a,b$ are arbitrary constants. Then the book concludes that y axis is the integral curve of the differential equation, but not the graph of the solution. I know the definition of the integral curve and the solution of an equation.
A detailed help would be much appreciated!
Thanks in advance!
$\endgroup$ 13 Answers
$\begingroup$Given an ODE, you have infinitely many integral curves corresponding to different initial conditions. Think of the initial condition as a free parameter.
Once you specify your initial condition, you have a solution for the ODE.
So you can think of the solution of the ODE one among the many integral curves that satisfies your IC.
For the reason above, some texts use solution curves instead of integral curves.
$\endgroup$ $\begingroup$A solution is an integral that is also continuous on the defined domain ($\mathbf{ Differential}$ $\mathbf{Equations}$ by Lomen&Lovelock). $y=kx$ on $(-\infty,0)\cup (0,\infty)$ is a solution because its derivative is just as required on the given domain ($(-\infty,0)\cup (0,\infty)$ and it is also continuous on this domain. But $x=0$ is not a solution because it is not defined on the domain hence not continuous there.
$\endgroup$ $\begingroup$The equation $\dot{x} = -y$, $\dot{y} = x$ has the general solution $(x, y) = A (\cos (t - \phi), \sin (t - \phi))$. Every solution of this equation lies on an integral curve $x^2 + y^2 = A^2$ which are circles. However these circles also admit parameterizations such as $(x,y) = A (\cos 2t, \sin 2t)$ which do not satisfy the ODE. Hence not every parametrization of an integral curve yields a solution but there is one parametrization of an integral curve which is a solution.
$\endgroup$
